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Multiplication and Division of Signed Numbers

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The basic operations with real numbers are presented in this chapter. The concept of absolute value is discussed both geometrically and symbolically. The geometric presentation offers a visual understanding of the meaning of |x|. The symbolic presentation includes a literal explanation of how to use the definition. Negative exponents are developed, using reciprocals and the rules of exponents the student has already learned. Scientific notation is also included, using unique and real-life examples. Objectives of this module: be able to multiply and divide signed numbers.

Overview

  • Multiplication of Signed Numbers
  • Division of Signed Numbers

Multiplication of Signed Numbers

Let us consider first the product of two positive numbers.

Multiply: 35 35 .
35 35 means 5+5+5=15 5+5+5=15 .

This suggests that

(positivenumber)(positivenumber)=positivenumber (positivenumber)(positivenumber)=positivenumber .

More briefly, (+)(+)=+ (+)(+)=+ .

Now consider the product of a positive number and a negative number.

Multiply: (3)(5) (3)(5) .
(3)(5) (3)(5) means (5)+(5)+(5)=15 (5)+(5)+(5)=15 .

This suggests that

(positivenumber)(negativenumber)=negativenumber (positivenumber)(negativenumber)=negativenumber

More briefly, (+)(-)=- (+)(-)=- .

By the commutative property of multiplication, we get

(negativenumber)(positivenumber)=negativenumber (negativenumber)(positivenumber)=negativenumber

More briefly, (-)(+)=- (-)(+)=- .

The sign of the product of two negative numbers can be determined using the following illustration: Multiply 2 2 by, respectively, 4,3,2,1,0,1,2,3,4 4,3,2,1,0,1,2,3,4 . Notice that when the multiplier decreases by 1, the product increases by 2.

4(2)=8 3(2)=6 2(2)=4 1(2)=2 } Asweknow,(+)()=. 0(2)=0 Asweknow,0(anynumber)=0. 4(2)=8 3(2)=6 2(2)=4 1(2)=2 } Asweknow,(+)()=. 0(2)=0 Asweknow,0(anynumber)=0.

1(2)=2 2(2)=4 3(2)=6 4(2)=8 } Thispatternsuggests()()=+. 1(2)=2 2(2)=4 3(2)=6 4(2)=8 } Thispatternsuggests()()=+.

We have the following rules for multiplying signed numbers.

Rules for Multiplying Signed Numbers

To multiply two real numbers that have

  1. the same sign, multiply their absolute values. The product is positive.
    (+)(+)=+ ()()=+ (+)(+)=+ ()()=+
  2. opposite signs, multiply their absolute values. The product is negative.
    (+)()= ()(+)= (+)()= ()(+)=

Sample Set A

Find the following products.

Example 1

86 86

Multiplytheseabsolutevalues. | 8 |=8 | 6 |=6 } 86=48 Sincethenumbershavethesamesign,theproductispositive. 86=+48 or86=48 Multiplytheseabsolutevalues. | 8 |=8 | 6 |=6 } 86=48 Sincethenumbershavethesamesign,theproductispositive. 86=+48 or86=48

Example 2

(8)(6) (8)(6)

Multiplytheseabsolutevalues. | 8 |=8 | 6 |=6 } 86=48 Sincethenumbershavethesamesign,theproductispositive. (8)(6)=+48 or(8)(6)=48 Multiplytheseabsolutevalues. | 8 |=8 | 6 |=6 } 86=48 Sincethenumbershavethesamesign,theproductispositive. (8)(6)=+48 or(8)(6)=48

Example 3

(4)(7) (4)(7)

Multiplytheseabsolutevalues. | 4 | =4 | 7 | =7 } 47=28 Sincethenumbershaveoppositesigns,theproductisnegative. (4)(7)=28 Multiplytheseabsolutevalues. | 4 | =4 | 7 | =7 } 47=28 Sincethenumbershaveoppositesigns,theproductisnegative. (4)(7)=28

Example 4

6(3) 6(3)

Multiplytheseabsolutevalues. | 6 | =6 | 3 | =3 } 63=18 Sincethenumbershaveoppositesigns,theproductisnegative. 6(3)=18 Multiplytheseabsolutevalues. | 6 | =6 | 3 | =3 } 63=18 Sincethenumbershaveoppositesigns,theproductisnegative. 6(3)=18

Practice Set A

Find the following products.

Exercise 1

3(8) 3(8)

Solution

24 24

Exercise 2

Exercise 3

(6)(5) (6)(5)

Solution

30

Exercise 4

(7)(2) (7)(2)

Solution

14

Exercise 5

(1)(4) (1)(4)

Solution

4 4

Exercise 6

(7)7 (7)7

Solution

49 49

Division of Signed Numbers

We can determine the sign pattern for division by relating division to multiplication. Division is defined in terms of multiplication in the following way.

If bc=a bc=a , then a b =c,b0 a b =c,b0 .

For example, since 34=12 34=12 , it follows that 12 3 =4 12 3 =4 .

Notice the pattern:

Since 34 bc=a =12 34 bc=a =12 , it follows that 12 3 a b =c =4 12 3 a b =c =4

The sign pattern for division follows from the sign pattern for multiplication.

  1. Since (+)(+) bc=a =+ (+)(+) bc=a =+ , it follows that (+) (+) a b =c =+ (+) (+) a b =c =+ , that is,

    (positivenumber) (positivenumber) =positivenumber (positivenumber) (positivenumber) =positivenumber

  2. Since ()() bc=a =+ ()() bc=a =+ , it follows that (+) () a b =c = (+) () a b =c = , that is,

    (positivenumber) (negativenumber) =negativenumber (positivenumber) (negativenumber) =negativenumber

  3. Since (+)() bc=a = (+)() bc=a = , it follows that () (+) a b =c = () (+) a b =c = , that is,

    (negativenumber) (positivenumber) =negativenumber (negativenumber) (positivenumber) =negativenumber

  4. Since ()(+) bc=a = ()(+) bc=a = , it follows that () () a b =c =+ () () a b =c =+ , that is

    (negativenumber) (negativenumber) =positivenumber (negativenumber) (negativenumber) =positivenumber

We have the following rules for dividing signed numbers.

Rules for Dividing Signed Numbers

To divide two real numbers that have

  1. the same sign, divide their absolute values. The quotient is positive.
    (+) (+) =+ () () =+ (+) (+) =+ () () =+
  2. opposite signs, divide their absolute values. The quotient is negative.
    () (+) = (+) () = () (+) = (+) () =

Sample Set B

Find the following quotients.

Example 5

10 2 10 2

| -10 |= 10 | 2 |= 2 } Dividetheseabsolutevalues. 10 2 =5 -10 2 =-5 Sincethenumbershaveoppositesigns,thequotientisnegative. | -10 |= 10 | 2 |= 2 } Dividetheseabsolutevalues. 10 2 =5 -10 2 =-5 Sincethenumbershaveoppositesigns,thequotientisnegative.

Example 6

35 7 35 7

| -35 |= 35 | -7 |= 7 } Dividetheseabsolutevalues. 35 7 =5 -35 -7 =5 Sincethenumbershavesamesigns,thequotientispositive. | -35 |= 35 | -7 |= 7 } Dividetheseabsolutevalues. 35 7 =5 -35 -7 =5 Sincethenumbershavesamesigns,thequotientispositive.

Example 7

18 9 18 9

| 18 |= 18 | -9 |= 9 } Dividetheseabsolutevalues. 18 9 =2 18 -9 =-2 Sincethenumbershaveoppositesigns,thequotientisnegative. | 18 |= 18 | -9 |= 9 } Dividetheseabsolutevalues. 18 9 =2 18 -9 =-2 Sincethenumbershaveoppositesigns,thequotientisnegative.

Practice Set B

Find the following quotients.

Exercise 7

24 6 24 6

Solution

4

Exercise 8

30 5 30 5

Solution

6 6

Exercise 9

54 27 54 27

Solution

2 2

Exercise 10

Sample Set C

Example 8

Find the value of 6(47)2(89) (4+1)+1 6(47)2(89) (4+1)+1 .

Using the order of operations and what we know about signed numbers, we get

6(47)2(89) (4+1)+1 = 6(3)2(1) (5)+1 = 18+2 5+1 = 20 4 = 5 6(47)2(89) (4+1)+1 = 6(3)2(1) (5)+1 = 18+2 5+1 = 20 4 = 5

Example 9

Find the value of z= xu s z= xu s if x=57,u=51,ands=2 x=57,u=51,ands=2 .

Substituting these values we get

z= 5751 2 = 6 2 =3 z= 5751 2 = 6 2 =3

Practice Set C

Exercise 11

Find the value of 7(48)+2(111) 5(16)17 7(48)+2(111) 5(16)17 .

Solution

1

Exercise 12

Find the value of P= n(n3) 2n P= n(n3) 2n , if n=5 n=5 .

Solution

1

Exercises

Find the value of each of the following expressions.

Exercise 13

(2)(8) (2)(8)

Solution

16

Exercise 14

(3)(9) (3)(9)

Exercise 15

(4)(8) (4)(8)

Solution

32

Exercise 16

(5)(2) (5)(2)

Exercise 17

(6)(9) (6)(9)

Solution

54

Exercise 18

(3)(11) (3)(11)

Exercise 19

(8)(4) (8)(4)

Solution

32

Exercise 20

(1)(6) (1)(6)

Exercise 21

(3)(12) (3)(12)

Solution

3636

Exercise 22

(4)(18) (4)(18)

Exercise 23

8(4) 8(4)

Solution

3232

Exercise 24

5(6) 5(6)

Exercise 25

9(2) 9(2)

Solution

1818

Exercise 26

7(8) 7(8)

Exercise 27

(6)4 (6)4

Solution

2424

Exercise 28

(7)6 (7)6

Exercise 29

(10)9 (10)9

Solution

9090

Exercise 30

(4)12 (4)12

Exercise 31

(10)(6) (10)(6)

Solution

6060

Exercise 32

(6)(4) (6)(4)

Exercise 33

(2)(6) (2)(6)

Solution

1212

Exercise 34

(8)(7) (8)(7)

Exercise 35

Exercise 36

42 6 42 6

Exercise 37

39 3 39 3

Solution

1313

Exercise 38

20 10 20 10

Exercise 39

45 5 45 5

Solution

9

Exercise 40

16 8 16 8

Exercise 41

25 5 25 5

Solution

55

Exercise 42

36 4 36 4

Exercise 43

8(3) 8(3)

Solution

11

Exercise 44

14(20) 14(20)

Exercise 45

20(8) 20(8)

Solution

28

Exercise 46

4(1) 4(1)

Exercise 47

04 04

Solution

44

Exercise 48

0(1) 0(1)

Exercise 49

6+17 6+17

Solution

1212

Exercise 50

151220 151220

Exercise 51

167+8 167+8

Solution

44

Exercise 52

2+710+2 2+710+2

Exercise 53

3(46) 3(46)

Solution

66

Exercise 54

8(512) 8(512)

Exercise 55

3(16) 3(16)

Solution

15

Exercise 56

8(412)+2 8(412)+2

Exercise 57

4(18)+3(103) 4(18)+3(103)

Solution

49

Exercise 58

9(02)+4(89)+0(3) 9(02)+4(89)+0(3)

Exercise 59

6(29)6(2+9)+4(11) 6(29)6(2+9)+4(11)

Solution

140140

Exercise 60

3(4+1)2(5) 2 3(4+1)2(5) 2

Exercise 61

4(8+1)3(2) 42 4(8+1)3(2) 42

Solution

77

Exercise 62

1(3+2)+5 1 1(3+2)+5 1

Exercise 63

3(42)+(3)(6) 4 3(42)+(3)(6) 4

Solution

33

Exercise 64

1(4+2) 1(4+2)

Exercise 65

1(61) 1(61)

Solution

55

Exercise 66

(8+21) (8+21)

Exercise 67

(821) (821)

Solution

13

Exercise 68

(106) (106)

Exercise 69

(52) (52)

Solution

33

Exercise 70

(711) (711)

Exercise 71

(812) (812)

Solution

4

Exercise 72

3[(1+6)(27)] 3[(1+6)(27)]

Exercise 73

2[(48)(511)] 2[(48)(511)]

Solution

44

Exercise 74

5[(1+5)+(68)] 5[(1+5)+(68)]

Exercise 75

[(49)+(28)] [(49)+(28)]

Solution

15

Exercise 76

3[2(15)3(2+6)] 3[2(15)3(2+6)]

Exercise 77

2[5(10+11)2(57)] 2[5(10+11)2(57)]

Solution

2

Exercise 78

P=RC. FindPifR=2000andC=2500. P=RC. FindPifR=2000andC=2500.

Exercise 79

z= xu s . Findzifx=23,u=25,ands=1. z= xu s . Findzifx=23,u=25,ands=1.

Solution

22

Exercise 80

z= xu s . Findzifx=410,u=430,ands=2.5. z= xu s . Findzifx=410,u=430,ands=2.5.

Exercise 81

m= 2s+1 T . Findmifs=8andT=5. m= 2s+1 T . Findmifs=8andT=5.

Solution

33

Exercise 82

m= 2s+1 T . Findmifs=10andT=5. m= 2s+1 T . Findmifs=10andT=5.

Exercise 83

Use a calculator. F=( p 1 p 2 ) r 4 9. FindFif p 1 =10, p 2 =8,r=3. F=( p 1 p 2 ) r 4 9. FindFif p 1 =10, p 2 =8,r=3.

Solution

1458

Exercise 84

Use a calculator. F=( p 1 p 2 ) r 4 9. FindFif p 1 =12, p 2 =7,r=2. F=( p 1 p 2 ) r 4 9. FindFif p 1 =12, p 2 =7,r=2.

Exercise 85

P=n(n1)(n2). FindPifn=4. P=n(n1)(n2). FindPifn=4.

Solution

120120

Exercise 86

P=n(n1)(n2)(n3). FindPifn=5. P=n(n1)(n2)(n3). FindPifn=5.

Exercise 87

P= n(n2)(n4) 2n . FindPifn=6. P= n(n2)(n4) 2n . FindPifn=6.

Solution

40

Exercises for Review

Exercise 88

((Reference)) What natural numbers can replace x x so that the statement 4<x3 4<x3 is true?

Exercise 89

((Reference)) Simplify (x+2y) 5 (3x1) 7 (x+2y) 3 (3x1) 6 (x+2y) 5 (3x1) 7 (x+2y) 3 (3x1) 6 .

Solution

( x+2y ) 2 ( 3x1 ) ( x+2y ) 2 ( 3x1 )

Exercise 90

((Reference)) Simplify ( x n y 3t ) 5 ( x n y 3t ) 5 .

Exercise 91

((Reference)) Find the sum. 6+(5) 6+(5) .

Solution

1111

Exercise 92

((Reference)) Find the difference. 2(8) 2(8) .

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